Mixing Problem need help work shown!

  • Thread starter johnq2k7
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  • #1
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Mixing Problem.... need help work shown!

A tank with a 1000 gallon capacity intially contains 500 gallons of water that is polluted with 50 lb of particulate matter. At time (t=0), pure water is added at a rate of 20 gal/min and the mixed solution is drained off at a rate of 10 gal/min. How much particulate matter is in the the tank when it reaches the point of overflowing?

work shown below:

Volume of Tank Capacity= 1000 gal
Intial Volume of Water= 500 gal
Intial Polluted Matter= 50 lb

Pure water Rate IN= 20 gal/min
Mixed Solution Rate OUT= 10 gal/min

Overflowing= excess of 1000 galloons

change in rate= change in rate in - change in rate out
= 20 gal/min - 10 gal/min
= 10 gal/min

change in rate therefore equals 10 gal/min

intial total amount of material in tank is : 500 gallons + 50 lbs
since 1 galloon equals approx. 8.34 lb

therefore, there is intially 917 galloons of material in the tank

how do i use this information that I processed to find the particulate matter once it reaches the point of overflowing which is (>=1000gal)

Please help me with this problem
 

Answers and Replies

  • #2
djeitnstine
Gold Member
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The rate of water flowing in is = rate in - rate out. or [tex]\frac{dA}{dt}= rate in - rate out[/tex]

oops I see you see that already. The rate you seem to be analyzing is that of the water. How about the particulate solution itself? How much is flowing in? How much is going out? So the rate of change of amount of solution must be negative because 0 - rate out. Also to examine the rate out. try using your units.

rate out is lb/min right? so... [tex]\frac{A(t) lb}{gal} \frac{gal}{min}[/tex] units will show you the answer.

where A(t) is amount in pounds (lb)
 
Last edited:
  • #3
djeitnstine
Gold Member
614
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Also I forgot to mention that you will need a separate DE to find out the time T it takes for the water to reach >1,000 gal then stick it into your particulate solution DE
 

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